# Quantum Entanglement States

In an earlier post, I described four simple thought experiments, involving some black boxes and two or more participants. As described there, the results of these experiments were inconsistent with any classical description, assuming that the boxes cannot communicate. However, I also stated that all of these experiments are consistent with quantum probability, and that I would give the mathematical details in a further post. I will do this now.

The standard constructions will be used, where the state ${\Psi}$ of a system is described by a unit vector in a Hilbert space ${\mathcal H}$, which will be finite dimensional in the examples here. An observable value of this system is described by a Hermitian operator A, so that ${A^*=A}$, and the probability of any specified (real) value ${a}$ being observed is

 $\displaystyle {\mathbb P}({\rm the\ observed\ value\ of\ }A{\rm\ is\ }a)=\lVert P\Psi\lVert^2,$

where P is the orthogonal projection on the eigenspace of A corresponding to eigenvalue a. In particular, the expected value of the observation is given by the inner product ${\langle\Psi,A\Psi\rangle}$.

We will be concerned with binary observables, such as whether a light glows red or green. These outcomes will be represented by the numbers 1 and -1, so that the corresponding Hermitian operator has only these eigenvalues. Equivalently, ${A^2=1}$ or, since it is Hermitian, A is unitary so that ${A^*A=1}$. In that case, the probability p of outcome 1 being observed, is related to the expected value by,

 \displaystyle \begin{aligned} &\langle\Psi,A\Psi\rangle=p.1+(1-p).(-1)=2p-1,\\ &p=\left(\langle\Psi,A\Psi\rangle+1\right)/2. \end{aligned} (1)

Operators A and B correspond to a simultaneously measurable pair of observables if they commute, ${AB=BA}$. Next, if we have two systems represented individually by Hilbert spaces ${\mathcal H_1}$ and ${\mathcal H_2}$, then the composite system is described by the tensor product ${\mathcal H_1\otimes\mathcal H_2}$. An observable on the first system represented by an operator A, is given on the composite system by the tensor product of operators ${A\otimes I}$. Similarly, observable operator B on the second system is represented by ${I\otimes B}$ on the composite one. The fact that these tensor products of operators commute corresponds to the fact that measurements on the separated components of a composite system do not interfere with each other, so are simultaneously measurable.

 $\displaystyle (A\otimes I)(I\otimes B)=A\otimes B=(I\otimes B)(A\otimes I).$

In our experiments, this means that the different boxes do not communicate.

#### Pauli Matrices

It will be convenient to make use of the Pauli spin matrices to represent the measurements associated with pressing a button on one of the black boxes. These are 2×2 complex matrices and, so, represent operators on a two dimensional Hilbert space, or qubit,

 $\displaystyle S_x=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},\ S_y=\begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix},\ S_z=\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.$

These are Hermitian, and can be used to represent the measurement of the spin of an electron, or other spin 1/2 particle, in a direction along the x, y and z axes respectively. More generally, for any unit vector ${v=(v_x,v_y,v_z)}$ in three dimensional Euclidean space, the matrix ${v_xS_x+v_yS_y+v_zS_z}$ represents a spin measurement in direction v. The following identities can be checked,

 \displaystyle \begin{aligned} & S_x^2=S_y^2=S_z^2=I,\\ & S_xS_y=-S_yS_x=iS_z,\\ & S_yS_z=-S_zS_y=iS_x,\\ & S_zS_x=-S_xS_z=iS_y. \end{aligned}

All of the eigenvalues of these matrices are equal to 1 or -1, representing the fact that the spin can only be in one of two states, up or down. The corresponding normalized eigenvectors are,

 ${\lvert\rightarrow\rangle = \frac{1}{\sqrt2}\begin{pmatrix}1 \cr 1\end{pmatrix}}$, ${\lvert\leftarrow\rangle = \frac{1}{\sqrt2}\begin{pmatrix}1 \cr -1\end{pmatrix}}$, ${\lvert\uparrow\rangle = \frac1{\sqrt2}\begin{pmatrix} 1 \cr i\end{pmatrix}}$, ${\lvert\downarrow\rangle = \frac1{\sqrt2}\begin{pmatrix} 1 \cr -i\end{pmatrix}}$, ${\lvert+\rangle = \begin{pmatrix} 1 \cr 0\end{pmatrix}}$, ${\lvert-\rangle = \begin{pmatrix} 0 \cr 1\end{pmatrix}}$,

The action of each of the Pauli matrices on each of the eigenvectors is given in the following table,

 ${S_x\lvert\rightarrow\rangle=\lvert\rightarrow\rangle}$, ${S_y\lvert\rightarrow\rangle=-i\lvert\leftarrow\rangle}$, ${S_z\lvert\rightarrow\rangle=\lvert\leftarrow\rangle}$, ${S_x\lvert\leftarrow\rangle=-\lvert\leftarrow\rangle}$, ${S_y\lvert\leftarrow\rangle=i\lvert\rightarrow\rangle}$, ${S_z\lvert\leftarrow\rangle=\lvert\rightarrow\rangle}$, ${S_x\lvert\uparrow\rangle=i\lvert\downarrow\rangle}$, ${S_y\lvert\uparrow\rangle=\lvert\uparrow\rangle}$, ${S_z\lvert\uparrow\rangle=\lvert\downarrow\rangle}$, ${S_x\lvert\downarrow\rangle=-i\lvert\uparrow\rangle}$, ${S_y\lvert\downarrow\rangle=-\lvert\downarrow\rangle}$, ${S_z\lvert\downarrow\rangle=\lvert\uparrow\rangle}$, ${S_x\lvert +\rangle=\lvert-\rangle}$, ${S_y\lvert +\rangle=i\lvert-\rangle}$, ${S_z\lvert +\rangle = \lvert+\rangle}$, ${S_x\lvert -\rangle=\lvert+\rangle}$, ${S_y\lvert -\rangle=-i\lvert+\rangle}$, ${S_z\lvert -\rangle = -\lvert-\rangle}$.

I list out all these identities now, so that it is easy to check the algebra in the examples below.

#### Experiment 1: GHZ Boxes

The first of our thought experiments concerned three participants, Alice, Bob and Charlie. They are each handed a black box having two buttons labelled L and R, and a single light which glows either red or green once a button is pressed. We set this up so that pressing a button triggers the measurement of a single qubit stored inside the box. Let us use ${S_z}$ for the measurement corresponding to button L and ${S_x}$ for that corresponding to R. The box will light up red if the outcome is 1, and green if it is -1. The combined states for the three boxes are then represented by the tensor product of the three two dimensional Hilbert spaces, and the observations of each of Alice, Bob and Charlie pressing either L and R are represented on this combined space by,

 \displaystyle \begin{aligned} A_L=S_z\otimes I\otimes I,\ & A_R=S_x\otimes I\otimes I,\\ B_L=I\otimes S_z\otimes I,\ & B_R=I\otimes S_x\otimes I,\\ C_L=I\otimes I\otimes S_z,\ & C_R=I\otimes I\otimes S_x. \end{aligned}

Although each of Alice’s observations commutes with each of Bob’s and Charlie’s, as must be the case if the boxes do not communicate, the different measurements that any individual participant can make do not commute with each other. In fact, they anti-commute,

 $\displaystyle A_LA_R=-A_RA_L,\ B_LB_R=-B_RB_L,\ C_LC_R=-C_RC_L.$

Recall the properties that the boxes were claimed to possess,

1. If Alice, Bob and Charlie all press L, then an even number of them (none or two) will see their box light up red.
2. If precisely one of Alice, Bob and Charlie press L, with the other two pressing R, then an odd number of them (one or three) will see their box light up red.

As explained in that post, these can be represented algebraically by the equations,

 \displaystyle \begin{aligned} &A_LB_LC_L=-1,\\ &A_LB_RC_R=A_RB_LC_R=A_RB_RC_L=1. \end{aligned} (2)

There, it was noted that the first line contradicts the second, if each of the variables take only the values 1 and -1. This implicitly made use of the rather obvious fact that if we multiply any two of these numbers together then the order does not matter. Multiplication of real numbers commutes. However, using the operators above, the anti-commutation properties cause a sign-change when the terms on the second line of (2) are multiplied,

 $\displaystyle (A_LB_RC_R)(A_RB_LC_R)(A_RB_RC_L)=-A_LB_LC_L.$

Then, (2) is no longer inconsistent and, in fact, the first equation is implied by the second line of equalities.

Actually, the observable operators defined here do not satisfy (2) on their own. This is not necessary for the properties listed above to hold though. It is only necessary that (2) holds when applied to the vector representing the combined state of the boxes. As is standard, we use ${\lvert ab\rangle}$ to represent a tensor product of vectors ${\lvert a\rangle}$ and ${\lvert b\rangle}$, and so on for the product of any number of vectors. In terms of the ‘spin up’ and ‘spin down’ states for the individual qubits, let us suppose that the system consisting of the three boxes has been prepared in a GHZ state,

 $\displaystyle \Psi=\frac1{\sqrt2}\left(\lvert\uparrow\uparrow\uparrow\rangle-\lvert\downarrow\downarrow\downarrow\rangle\right).$ (3)

Using the identities listed above for applying the Pauli spin matrices to the spin up and spin down states,

 \displaystyle \begin{aligned} S_z\otimes S_z\otimes S_z \Psi &= -\Psi \\ S_z\otimes S_x\otimes S_x \Psi &= S_x\otimes S_z\otimes S_x \Psi \\ &= S_x\otimes S_x\otimes S_z \Psi \\ &= \Psi. \end{aligned}

Hence, equations (2) are satisfied when applied to ${\Psi}$, so the boxes do indeed satisfy the claimed properties.

There is one remaining thing to be proved for this experiment. Recall that I claimed that the boxes have the following statistical behaviour,

• If precisely two of Alice, Bob and Charlie press L, or none of them do, then the guaranteed properties are not applicable. Each box lights up red or green with probability one half, and the boxes are independent of each other. That is, each of the 8 possible combinations of results occur with probability 1/8.
• If precisely one of Alice, Bob and Charlie press L, or if they all do, then the guaranteed properties apply. In this case, each of the 4 possible combinations of results obeying the required parity constraint occurs with probability 1/4.

We could prove these properties in a systematic fashion by computing the size of the projections of the GHZ state (3) onto each of the combined eigenstates of the observable operators. Alternatively, a bit of symmetry can be used.

As ${S_y}$ anticommutes with each of the other two Pauli matrices, the effect of applying it to any any state is to flip the measured values of ${S_x}$ and ${S_z}$. Also, using the identities above for Pauli matrices,

 \displaystyle \begin{aligned} I\otimes S_y\otimes S_y \Psi &= S_y\otimes I\otimes S_y \Psi \\ &= S_y\otimes S_y\otimes I \Psi \\ &= \Psi. \end{aligned}

This shows that flipping the results of any two of the observers (i.e., changing red lights to green, and vice-versa) does not affect the state, so the distribution is left unchanged. Similarly, ${S_z}$ anticommutes with ${S_x}$ and commutes with itself. So, the effect of applying ${S_z\otimes S_z\otimes S_z}$ to the state is to flip the result for each of the three observers who press R. However, we have already shown that this operator only changes the sign of ${\Psi}$, which does not affect any probabilities. So, whichever measurements each of the three observers makes, we know the distribution of outcomes is unchanged under the following operations.

• Flipping the result of any fixed pair of participants.
• Flipping the result of each participant who presses R.

As these operations, or combinations of them, can change any of the allowed outcomes for a specific choice of buttons to any of the others, each of these must have the same probability.

#### Experiment 2: Mermin’s Boxes

Moving on to the second experiment, there are now two participants, Alice and Bob, who are each given a simple black box with two buttons labelled L and R, and a single light which glows either red or green when a button is pressed. Their claimed properties are,

1. If Alice and Bob both press L, then at least one of the boxes lights up green.
2. If one of Alice and Bob presses L and the other presses R, then at least one of the boxes lights up red.
3. If they both press R, then the boxes sometimes both light up green.

For this experiment, the boxes can be constructed in the same way as in the first experiment. Pressing L causes the measurement represented by ${S_z}$ to be performed, and pressing R causes ${S_x}$ to be measured. So, on the combined state for the two boxes, the observables associated with Alice and Bob pressing L or R are,

 \displaystyle \begin{aligned} A_L=S_z\otimes I,\ & A_R=S_x\otimes I,\\ B_L=I\otimes S_z,\ & B_R=I\otimes S_x. \end{aligned} (4)

The light glows red if the outcome is 1, and green if it is -1. In terms of the eigenvectors described above, suppose that the boxes are prepared in the combined state,

 $\displaystyle \Psi=\frac1{\sqrt3}\left(\lvert+-\rangle+\lvert-+\rangle+\lvert--\rangle\right).$

Noting that ${\lvert+\rangle}$ and ${\lvert-\rangle}$ are the eigenvectors of ${S_z}$, with eigenvalues 1 and -1 respectively, we see that in this state, if both Alice and Bob press L, then they see the results red/green, green/red and green/green, each with equal probability. In particular, at least one sees their box light green as claimed in the first property.

Next, suppose that one of the participants presses L and the other presses R. From the properties listed above for Pauli matrices, ${(1-S_z)\lvert+\rangle}$ and ${(1-S_x)(\lvert+\rangle+\lvert-\rangle)}$ are both equal to zero, and we obtain,

 \displaystyle \begin{aligned} &(1-A_L)(1-B_R)\Psi=(I-S_z)\otimes(I-S_x)\Psi=0,\\ &(1-A_R)(1-B_L)\Psi=(I-S_x)\otimes(I-S_z)\Psi=0. \end{aligned}

Hence, at least one of the participants measures the result 1, and their light glows red, as claimed by the second property.

It remains to show that their boxes sometimes both light up green when they both press R. I will actually show the full statistical properties for all combinations of button presses, which was described in the original post by the following table.

The first row was already described above, so only the remaining three are left. As the probabilities of Alice’s outcomes does not depend on which button Bob presses, we see that if she presses L, then she sees green with probability 2/3. Similarly for Bob. We can also compute the probability that Alice (or Bob) sees their box light green if they press R by projecting onto the eigenspace corresponding to ${A_R}$ taking the value -1,

 \displaystyle \begin{aligned} \frac12(1-A_R)\Psi &=\frac12(1-S_x)\otimes I\Psi\\ &=\frac1{2\sqrt3}\left(\lvert-+\rangle-\lvert++\rangle\right). \end{aligned}

Taking the inner product with ${\Psi}$, we obtain a probability of 1/6.

Look at the second row of the table, where Alice presses L and Bob presses R. The only way that Alice can see green, which we argued has probability 2/3, is if Bob’s box lights red. The only way that Bob can see green, which we showed has probability 1/6, is if Alice’s box lights red. As the row must sum to 1, the remaining ‘RR’ entry must have probability 1/6. The third row of the table follows in the same way as for the second, by exchanging the roles of Alice and Bob.

For the final row of the table, where Alice and Bob both press R, we have already argued that they each have probability 1/6 of their light glowing green. This means that the first two entries sum to 1/6, as do the first and third. Once we have computed the probability for them both seeing green, this completes the table. Projecting onto the joint eigenspace for ${A_R}$ and ${B_R}$ taking the values -1,

 \displaystyle \begin{aligned} \frac14(1-A_R)(1-B_R)\Psi &=(1-S_x)\otimes(1-S_x)\Psi\\ &=\frac1{4\sqrt3}\left(\lvert+-\rangle+\lvert-+\rangle-\lvert--\rangle-\lvert++\rangle\right). \end{aligned}

Taking the inner product with ${\Psi}$ gives 1/12 as required.

#### Experiment 3: Bell’s Boxes

Turning to experiment 3 — or, at least, the modified version which is actually possible — we again have the two participants Alice and Bob, each of whom is given a simple black box having two buttons, labelled L and R, and a light which glows either green or red when one of the buttons is pressed. The following properties are satisfied with probability ${(2+\sqrt2)/4}$, which is about 85%.

1. If at least one of Alice or Bob presses L, then they see the same result. Either, both their boxes light red or both light green.
2. If both of Alice and Bob press R, then they see different results. One of their boxes lights red and one lights green.

As was explained, according to classical logic this is impossible if the boxes do not communicate. Applying quantum theory, we can reuse the boxes from the first two experiments. Each box contains a qubit, pressing L causes a measurement given by the operator ${S_z}$, and pressing R is given by ${S_x}$. The light glows red if the observed value is 1, and green if it is -1. The observables corresponding to each of the button presses that Alice and Bob can make are again given by (4),

 \displaystyle \begin{aligned} A_L=S_z\otimes I,\ & A_R=S_x\otimes I,\\ B_L=I\otimes S_z,\ & B_R=I\otimes S_x. \end{aligned}

The only thing which distinguishes this experiment from the previous one is that we start with a different quantum state for the pair of qubits.

 $\displaystyle \Psi=\frac1{\sqrt2}\left(\lvert\uparrow\uparrow\rangle+\omega\lvert\downarrow\downarrow\rangle\right),$

where ${\omega=\sqrt i=(1+i)/\sqrt2}$. If you are familiar with Bell states, you may wonder why I have a factor ${\omega}$ which is not usually present. The reason for it is that in descriptions of Bell tests, one measurement apparatus is usually rotated by 45 degrees from the other (or, 22.5 degrees when using photons). I prefer not to do this here, and instead apply the rotation to the state. This enables the boxes to be identical, and the same as those used in the first two experiments.

Applying the operators for each of the possible pairs of measurements for Alice and Bob,

 \displaystyle \begin{aligned} &A_LB_L\Psi=S_z\otimes S_z\Psi = \frac1{\sqrt2}\left(\omega\lvert\uparrow\uparrow\rangle+\lvert\downarrow\downarrow\rangle\right),\\ &A_RB_R\Psi=S_x\otimes S_x\Psi = \frac{-1}{\sqrt2}\left(\omega\lvert\uparrow\uparrow\rangle+\lvert\downarrow\downarrow\rangle\right),\\ &A_LB_R\Psi=S_z\otimes S_x\Psi = \frac{-i}{\sqrt2}\left(\omega\lvert\uparrow\uparrow\rangle-\lvert\downarrow\downarrow\rangle\right),\\ &A_RB_L\Psi=S_x\otimes S_z\Psi = \frac{-i}{\sqrt2}\left(\omega\lvert\uparrow\uparrow\rangle-\lvert\downarrow\downarrow\rangle\right). \end{aligned}

The expected value of these operators is given by the inner products,

 \displaystyle \begin{aligned} &\langle\Psi,A_LB_L\Psi\rangle=-\langle\Psi,A_RB_R\Psi\rangle=\frac12(\bar\omega+\omega)=\frac{\sqrt2}2,\\ &\langle\Psi,A_LB_R\Psi\rangle=\langle\Psi,A_RB_L\Psi\rangle=\frac i2(\bar\omega-\omega)=\frac{\sqrt2}2. \end{aligned}

Recalling equation (1) for the probability p that any particular one of the operators

 $\displaystyle A_LB_L,\ A_LB_R,\ A_RB_L,\ -A_RB_R,$

gives the value 1 when measured, gives,

 $\displaystyle p=\left(\sqrt2/2+1\right)/2.$

This is equivalent to the claimed properties of the boxes.

Finally, using the properties of Pauli matrices listed above, note that the state is unchanged under an application of ${S_y\otimes S_y}$. However, this operator anti-commutes with each of the observables ${A_L,A_R,B_L,B_R}$, so has the effect of flipping the colours of both lights. That is, inverting the light colours so that red becomes green and vice-versa, is a symmetry of the quantum state and does not change the statistical behaviour of the boxes.

#### Experiment 4: Magic Squares

Finally, we look at the fourth experiment. This again involved two participants, each handed a black box. In this case, though, the boxes have three buttons labelled 1, 2 and 3, and three lights, one above each button. It was claimed that they satisfy the following properties.

1. When one of the buttons on a box is pressed, its lights instantly turn on, each either green or red, such that the number of red lights is even (zero or two).
2. If Alice presses button i and Bob presses button j, then Alice’s j’th light and Bob’s i’th light are different colours.

As explained in the original post, without any communication occurring between the boxes, this behaviour is not allowed by classical logic. We related this to the impossibility of a certain kind of 3×3 magic square. Slightly updating the labelling scheme for the cells, to fit in more conveniently with the current construction, it is impossible to fill in each cell with either 1 or -1, such that each row has product equal to -1 and each column has product 1. That is, we cannot have an even number of 1’s in each row and an even number of -1’s in each column. We can try, as in the attempt below but, once we get to the final cell, its value will have to be chosen differently according to whether we want to satisfy the constraint for its row, or the constraint for its column.

 1 -1 1 -1 -1 -1 -1 1 ?

However, consider filling in the entries with operators instead, as in the below Mermin-Peres magic square. Using the notation above for the Pauli matrices, and I for the identity operator on a single qubit:

 ${I\otimes S_z}$ ${S_x\otimes I}$ ${-S_x\otimes S_z}$ ${S_z\otimes I}$ ${I\otimes S_x}$ ${-S_z\otimes S_x}$ ${S_z\otimes S_z}$ ${S_x\otimes S_x}$ ${S_y\otimes S_y}$

Its properties are easily established using the identities above for Pauli matrices and, as suggested by the name, they do seem quite magical. First, each operator is Hermitian and has square equal to ${I_2\equiv I\otimes I}$, so only has eigenvalues 1 and -1. Next, all three operators in any specified row or column commute with each other, so represent simultaneously observable values. Finally, the product of all three elements of any row gives ${-I_2}$ and the product of the elements of any column gives ${I_2}$. This is a quantum version of the classically impossible magic square described above. The reason why such a construction is possible, is that the operators do not all commute with each other, so it is not possible to simultaneously observe the value of all entries of the square. Actually, any pair of entries which do not lie in the same row or column will anti-commute.

The Mermin-Peres magic square can be used to construct boxes with the properties claimed above. First, let ${M_{ij}}$ denote the operator in row i and column j. We will suppose that each box contains a pair of qubits, and that pressing button i triggers a measurement of the three commuting observables ${M_{i1},M_{i2},M_{i3}}$. Then, the j’th light glows red if ${M_{ij}}$ is measured to be 1, and green if it is -1. As the product of each row of the Mermin-Peres magic square equals -1, this setup guarantees that each box will show an even number of red lights regardless of which button is pressed, and regardless of the initial quantum state. So, the first property is guaranteed.

The way that I have constructed the boxes, they both measure rows of the magic square, rather than columns. This was only done so that, as in the previous experiments, the two boxes are equivalent. However, it is easy to exchange rows and columns with the unitary operator,

 $\displaystyle U=\frac1{\sqrt2}(S_x-S_z)\otimes S_y.$

This represents a 180 degree rotation of the first qubit about the axis ${(1,0,-1)}$, and a 180 degree rotation of the second qubit about the y axis. It also exchanges the rows and columns of the Mermin-Peres magic square, together with a sign change,

 $\displaystyle U^*M_{ij}U=-M_{ji}.$

So, the boxes can also be considered to be measuring the columns of the magic square, after applying this operator to the state and flipping the colours of the lights.

On the combined state containing each of Alice and Bob’s boxes, their respective measurements are represented by the tensor product operators,

 \displaystyle \begin{aligned} &A_{ij}=M_{ij}\otimes I_2,\\ &B_{ij}=I_2\otimes M_{ij}. \end{aligned}

Here, ${A_{ij}}$ represents the colour of Alice’s j’th light if she presses i, and similarly ${B_{ij}}$ represents the color of Bob’s lights, with 1 corresponding to red and -1 to green. Consider the combined quantum state,

 $\displaystyle \Psi = \frac12\left(\lvert++++\rangle+\lvert+-+-\rangle+\lvert-+-+\rangle+\lvert----\rangle\right).$

Applying the properties of Pauli matrices above, the following can be checked,

 $\displaystyle A_{ij}B_{ij}\Psi=M_{ij}\otimes M_{ij}\Psi=\Psi.$

If they are in such a state, then Alice and Bob’s boxes will always give the same result if they both press the same numbered button. This is not what we want, so consider applying the rotation U to Bob’s qubits,

 $\displaystyle \Phi=I_2\otimes U\Psi.$

Using the fact that U exchanges rows and columns of the Mermin-Peres magic square and changes the sign,

 \displaystyle \begin{aligned} A_{ij}B_{ji}\Phi &= M_{ij}\otimes(M_{ji}U)\Psi\\ &=-(I_2\otimes U)(M_{ij}\otimes M_{ij})\Psi\\ &=-\Phi. \end{aligned}

If the boxes are initially in state ${\Phi}$, then ${A_{ij}B_{ji}}$ has the value -1, meaning that if Alice presses button i and Bob presses j, then her j’th light and Bob’s i’th light will be different colours. This is the second claimed property.

It remains to prove the following complete statistical description of the boxes’ behaviour.

• For any choice of buttons by Alice and Bob, the boxes light up completely at random, subject to the two guaranteed properties. There are a total of eight possible combinations of ways in which the lights can glow, each of which will occur with probability 1/8.

Consider the case where Alice presses button i and Bob presses j. Choosing any ${i^\prime\not=i}$ and ${j^\prime\not=j}$, we have already noted that the unitary operator

 $\displaystyle C\equiv -A_{i^\prime j^\prime}B_{j^\prime i^\prime}$

preserves the state ${\Phi}$. Also, as each pair of entries of the Mermin-Peres magic square anticommute, if they do not lie in a common row or column, we see that C anticommutes with ${A_{ik}}$ for each ${k\not=j^\prime}$ and with ${B_{jk}}$ for each ${k\not=i^\prime}$. Its effect is then to flip the colours of the corresponding lights on each of the boxes. That is, if we are to flip the colours of two of Alice’s lights, including light number j, and two of Bob’s including light number i, then this has no effect on the probabilities of each possible outcome. As every possible outcome satisfying the two required properties can be transformed into every other possible outcome by such operations, we conclude that they all occur with the same probability.

We are now done with all of the quantum mechanical descriptions for the thought experiments. Before finishing, I briefly note how the state in the final experiment could be constructed. We start with a Bell state for a pair of qubits, such as a pair of spin 1/2 particles.

 $\displaystyle \varphi=\frac1{\sqrt2}\left(\lvert++\rangle+\lvert--\rangle\right).$

This is as used in many Bell test experiments. Next, we consider preparing two pairs of particles independently, each in the Bell state above. The combined state is,

 $\displaystyle \varphi\otimes\varphi=\frac12\left(\lvert++++\rangle+\lvert++--\rangle+\lvert--++\rangle+\lvert----\rangle\right).$

This is similar to ${\Psi}$ above, except that the second and third qubits have been exchanged. This means that if we independently prepare two pairs of qubits, each in a Bell state, and put one qubit from each pair in Alice’s box and the other in Bob’s box, then the boxes will be in state ${\Psi}$. Finally, apply the rotation U defined above to Bob’s qubits to obtain the required state ${\Phi}$.